The molar conductivity of $0.025 \ M$ methanoic acid is $46.1 \ S \ cm^2 \ mol^{-1}$. Calculate its degree of dissociation. Given: $\lambda^0(H^{+}) = 349.6 \ S \ cm^2 \ mol^{-1}$ and $\lambda^0(HCOO^{-}) = 54.6 \ S \ cm^2 \ mol^{-1}$.

Given $l/a = 0.5 \ cm^{-1}$,$R = 50 \ \Omega$,$N = 1.0$. The equivalent conductance of the electrolytic cell is ................ $\Omega^{-1} \ cm^2 \ gm \ eq^{-1}$.

The molar conductivity of $0.4 \ M \ KCl$ solution is $2.5 \times 10^2 \ \Omega^{-1} \ cm^2 \ mol^{-1}$. What is the resistivity of the solution?

If $l = \text{length}$,$R = \text{Resistance}$,and $A = \text{Area of cross-section}$,then which of the following relations is correct?

Resistance of a conductivity cell at $298 \ K$ for $0.0100 \ M \ KCl$ is $161.8 \ \Omega$. Resistance becomes $190 \ \Omega$ when $0.005 \ M \ NaOH$ solution is filled in this cell. Calculate: $(i)$ cell constant,$(ii)$ specific conductivity for $NaOH$ solution,and $(iii)$ molar conductivity.

The resistance of $0.01 \ M$ $KCl$ solution at $298 \ K$ is $1500 \ \Omega$. If the conductivity of $0.01 \ M$ $KCl$ solution at $298 \ K$ is $0.1466 \times 10^{-3} \ S \ cm^{-1}$,the cell constant of the conductivity cell in $cm^{-1}$ is: